1) Sort all the elements of the array.
2) Find the index of first positive element, this is initial value of right index r.
3) Initialize: left index l = r – 1. min_sum = INT_MIN
4) In a loop, look for the candidates by comparing sum with minimum sum. If arr[l] + arr[r] becomes negative then increment the right index r, else decrement left index l.
#include
#include
#include
void quickSort(int *, int, int);
/* Function to print pair of elements having minimum sum */
void minAbsSumPair(int arr[], int arr_size)
{
int l, r = 0, min_sum, sum = 0, min_l, min_r;
/* Array should have at least two elements*/
if(arr_size < 2)
{
printf("Invalid Input");
return;
}
/* Sort the elements */
quickSort(arr, 0, arr_size-1);
/* Find the first positive element. Note that we have condition "r < arr_size -1"
-1 is there to handle the cases when all elements are negative */
while(arr[r] < 0 && r < arr_size - 1)
r++;
/* If all elements are positive then first two elements
are the minimum sum elements */
if(r == 0)
{
printf(" The first two elements %d and %d have minimum sum",
arr[0], arr[1]);
return;
}
/* Start looking for the pair from the first positive element
and last negative element */
l = r - 1;
min_sum = arr[l] + arr[r];
min_l = l; /* min_l is for the left element of minimum sum pair*/
min_r = r; /* min_r is for the right element of minimum sum pair*/
while(l >= 0 && r < arr_size)
{
sum = arr[l] + arr[r];
/*If abs(sum) is less then update the result items*/
if(abs(sum) < abs(min_sum))
{
min_sum = sum;
min_l = l;
min_r = r;
}
if(sum < 0)
r++;
else
l--;
printf (" %d %d %d \n ",min_sum,l,r);
}
printf(" The two elements whose sum is minimum are %d and %d",
arr[min_l], arr[min_r]);
}
/* Driver program to test above function */
int main()
{
int arr[] = {1, 60, -10, 70, -80, 85};
minAbsSumPair(arr, 6);
getchar();
return 0;
}
/* FOLLOWING FUNCTIONS ARE ONLY FOR SORTING
PURPOSE */
void exchange(int *a, int *b)
{
int temp;
temp = *a;
*a = *b;
*b = temp;
}
int partition(int arr[], int si, int ei)
{
int x = arr[ei];
int i = (si - 1);
int j;
for (j = si; j <= ei - 1; j++)
{
if(arr[j] <= x)
{
i++;
exchange(&arr[i], &arr[j]);
}
}
exchange (&arr[i + 1], &arr[ei]);
return (i + 1);
}
/* Implementation of Quick Sort
arr[] --> Array to be sorted
si --> Starting index
ei --> Ending index
*/
void quickSort(int arr[], int si, int ei)
{
int pi; /* Partitioning index */
if(si < ei)
{
pi = partition(arr, si, ei);
quickSort(arr, si, pi - 1);
quickSort(arr, pi + 1, ei);
}
}
Tim Complexity O(nlogn)
Space Complexity O(1)
Run Here https://ideone.com/Y4sk0
4 comments :
how is the TC O(nlogn) ?
can u explain?
thanks:)
@anonymous dude Sorting using Quick Sort Takes O(nlogn) time using Divide & Conquer Algo. its recurrence relation is given by
T(N)=2*T(n/2)+O(N)
See Master Theorem for details do notify me if need more clarification
Happy Coding :)
Shashank
the quicksort algo that u have used divides the list into lists having (n-1) elements and 0 elements.in that case worst case is o(n2) not o(nlogn)
The algorithm does not work with array {-1000, 1, 2}.
The output is 998 = abs(-1000+2)
But expected value 3 = abs(1,2)
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